Some Useful Trigonometric Limits

Before I discuss differentiating trigonometric functions, we should probably review a few key useful limits.

0. Useful Statements. We will use some statements that may or may not be obvious.

Squeeze Theorem. Suppose we have $f(x)\leq g(x)\leq h(x)$ for all $x\in (a,c)$. If $b\in(a,c)$ and
$\displaystyle L = \lim_{x\to b}f(x)=\lim_{x\to b}h(x)$
then
$\displaystyle \lim_{x\to b}g(x) = L$.

1. Sinc Limit. We will consider a function (called the “sinc” function) $\mathrm{sinc}(x)=x^{-1}\sin(x)$. What happens as $x\to0$? Lets think about it.

We will consider $0, then we know
(1)$\displaystyle \sin(x)\leq x\leq\tan(x)$.
Divide through by $\sin(x)$
(2)$\displaystyle 1\leq \frac{x}{\sin(x)}\leq\frac{\tan(x)}{\sin(x)}=\frac{1}{\cos(x)}$.
Well, we know that
(3)$\displaystyle \lim_{x\to0}\frac{1}{\cos(x)}=1$.
Thus we have our limit bounded
(4)$\displaystyle \lim_{x\to 0}1\leq\lim_{x\to0}\frac{x}{\sin(x)}\leq\lim_{x\to0}\frac{1}{\cos(x)}=1$.
We rewrite this to stress the boundedness:
(4′)$\displaystyle 1\leq\lim_{x\to0}\frac{x}{\sin(x)}\leq1$.
This implies
(5)$\displaystyle \lim_{x\to0}\frac{x}{\sin(x)}=1$.
More importantly, this implies (taking the recipricol):
(6)$\displaystyle \lim_{x\to0}\frac{\sin(x)}{x}=1$.
This is one important limit.

2. Cosine Version. This is all very nice, but lets consider the limit
(7)$\displaystyle \lim_{x\to0}\frac{1-\cos(x)}{x}=\textbf{??}$
What to do? Well, we can do the following trick
(8)$\displaystyle \lim_{x\to0}\frac{1-\cos(x)}{x}=\lim_{x\to0}\frac{1-\cos(x)}{x}\frac{1+\cos(x)}{1+\cos(x)}$
The numerator (top of the fraction) becomes
$\displaystyle (1-\cos(x))(1+\cos(x)) = 1 - \cos^{2}(x)=\sin^{2}(x)$.
We plug this in, rewriting our limit as
(9)$\displaystyle \lim_{x\to0}\frac{1-\cos(x)}{x}=\lim_{x\to0}\frac{\sin^{2}(x)}{x(1+\cos(x))}$.
What to do now?

We use the squeeze theorem, noting that the fraction has bounds
(10)$\displaystyle \frac{\sin^{2}(x)}{2x}\leq\frac{\sin^{2}(x)}{x(1+\cos(x))}\leq\frac{\sin^{2}(x)}{x}$.
Using limit properties, we find that
(11)$\displaystyle \lim_{x\to0}\frac{\sin^{2}(x)}{2x}=\frac{1}{2}\lim_{x\to0}\frac{\sin(x)}{x}\lim_{x\to0}\sin(x)=(1)\cdot(0)=0$.
Thus our problem has bounds
(12)$\displaystyle 0\leq\lim_{x\to0}\frac{\sin^{2}(x)}{x(1+\cos(x))}\leq0$.
The squeeze theorem produces the solution
(13)$\displaystyle \lim_{x\to0}\frac{\sin^{2}(x)}{x(1+\cos(x))}=0$.
Rewriting the function in familiar notation, we have
(13′)$\displaystyle \lim_{x\to0}\frac{1-\cos(x)}{x}=0$.

3. Punchline. So, we have just proven two important limits:
(14a)$\displaystyle \lim_{x\to0}\frac{\sin(x)}{x}=1$.
(14b)$\displaystyle \lim_{x\to0}\frac{1-\cos(x)}{x}=0$.

Exercise 1. Prove or find a counter-example: for any fixed constant $c$ we have $\displaystyle\lim_{x\to\infty} x\sin(c/x) = c$.